A 25-mL sample of 0.160M solution of NaOH is titrated with 17 mL of an unknown solution of H2SO4. What is the molarity of the sulfuric acid solution?

A. 0.004M H2SO4
B. 0.235M H2SO4
C. 0.117M H2SO4
D. 0.002M H2SO4

1. Write and balance the equation.

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

2. mols NaOH = M x L = ?

3. Using the coefficients in the balanced equation, convert mols NaOH to mols H2SO4.
? mols NaOH x (1 mol H2SO4/2 mols H2SO4) = ? mols H2SO4.

4. Now convert mols H2SO4 and volume to M.
M = mols H2SO4/L H2SO4

0.117M H2

SO4

To find the molarity of the sulfuric acid solution, we can use the formula:

M1V1 = M2V2

Where:
M1 = molarity of NaOH solution
V1 = volume of NaOH solution
M2 = molarity of H2SO4 solution
V2 = volume of H2SO4 solution

Given:
M1 = 0.160M
V1 = 25 mL = 0.025 L
V2 = 17 mL = 0.017 L

Substituting the given values into the formula:

(0.160M)(0.025 L) = M2(0.017 L)

Solving for M2:

(0.004) = M2(0.017)

M2 = 0.004/0.017

M2 ≈ 0.235

Therefore, the molarity of the sulfuric acid solution is approximately 0.235M.

So, the correct answer is B. 0.235M H2SO4.

To determine the molarity of the sulfuric acid solution, you can use the concept of stoichiometry and the balanced chemical equation of the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4).

The balanced chemical equation is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the equation, you can see that the ratio between NaOH and H2SO4 is 2:1.

Given that you have a 25 mL sample of a 0.160M NaOH solution, you can use the formula:
M1V1 = M2V2

Where:
M1 = initial molarity of NaOH
V1 = initial volume of NaOH
M2 = final molarity of H2SO4
V2 = final volume of H2SO4

Plugging in the values:
M1 = 0.160 M
V1 = 25 mL
M2 = ?
V2 = 17 mL

Rearranging the formula to find M2:
M2 = (M1 * V1) / V2

Now, substitute the given values:
M2 = (0.160 M * 25 mL) / 17 mL

Calculating the result:
M2 = 0.235 M

Therefore, the molarity of the sulfuric acid solution is 0.235M H2SO4. So, the correct answer is option B.